There seems to be one character missing. But I gather that <em>f(x)</em> needs to satisfy
•
divides 
•
divides 
I'll also assume <em>f(x)</em> is monic, meaning the coefficient of the leading term is 1, or

Since
divides
, and

where
is degree-2, and we can write it as

Now, we have

so if
divides
, then
is degree-2, so
is degree-4, and we can write

where
is degree-1.
Expanding the left side gives

and dividing by
leaves no remainder. If we actually compute the quotient, we wind up with

If the remainder is supposed to be zero, then

Adding these equations together and grouping terms, we get

Then
, and you can solve for <em>a</em> and <em>b</em> by substituting this into any of the three equations above. For instance,

So, we end up with

A = amount invested at 4%.
b = amount invested at 9%.
we know the total amount invested was $26500, thus
a + b = 26500.
whatever% of anything is just (whatever/100) * anything.
how much is 4% of a? well, is just (4/100) * a, or
0.04a.
how much is 9% of b? well, is just (9/100) * b, or
0.09b.
we know the interest yielded for both amounts adds up to $1510, thus
0.04a + 0.09b = 1510.

how much was invested at 9%? well, b = 26500 - a.
Answer:
5i√10
Step-by-step explanation:
hope this answers your question
:)
Answer:
1. the range of f^-1(x) is {10, 20, 30}.
2. the graph of f^-1(x) will include the point (0, 3)
3. n = 8
Step-by-step explanation:
1. The domain of a function is the range of its inverse, and vice versa. The range of f^-1(x) is {10, 20, 30}.
__
2. See above. The domain and range are swapped between a function and its inverse. That means function point (3, 0) will correspond to inverse function point (0, 3).
__
3. The n-th term of an arithmetic sequence is given by ...
an = a1 +d(n -1)
You are given a1 = 2, a12 = 211, so ...
211 = 2 + d(12 -1)
209/11 = d = 19 . . . . . solve the above equation for the common difference
Now, we can use the same equation to find n for an = 135.
135 = 2 + 19(n -1)
133/19 = n -1 . . . . . . . subtract 2, divide by 19
7 +1 = n = 8 . . . . . . . . add 1
135 is the 8th term of the sequence.
Answer: B 163 degrees
Step-by-step explanation:
That's an obtuse angle, bigger than 90 degrees, so we read the appropriate upper scale for the angle.